Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 83253 by mathmax by abdo last updated on 29/Feb/20

calculate ∫_0 ^∞ (dx/((x^4 −x^2 +1)^2 ))

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Commented by abdomathmax last updated on 01/Mar/20

sorry the Q is calculate ∫_(−∞) ^(+∞) (dx/((x^4 −x^2 +1)^2 ))

$${sorry}\:{the}\:{Q}\:{is}\:{calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Commented by mathmax by abdo last updated on 01/Mar/20

let I =∫_(−∞) ^(+∞) (dx/((x^4 −x^2 +1)^2 )) let ϕ(z)=(1/((z^4 −z^2 +1)^2 )) poles of ϕ? z^4 −z^2 +1=0⇒t^2 −t+1=0 (t=z^2 ) Δ=1−4=−3 ⇒t_1 =((1+i(√3))/2) =e^((iπ)/3) and t_2 =((1−i(√3))/2)=e^(−((iπ)/3)) t^2 −t+1 =(t−e^((iπ)/3) )(t−e^(−((iπ)/3)) ) ⇒z^4 −z^2 +1 =(z^2 −e^((iπ)/3) )(z^2 −e^((−iπ)/3) ) =(z−e^((iπ)/6) )(z+e^((iπ)/6) )(z−e^(−((iπ)/6)) )(z+e^((−iπ)/6) )⇒ ϕ(z)=(1/((z−e^((iπ)/6) )^2 (z+e^((iπ)/6) )^2 (z−e^(−((iπ)/6)) )^2 (z+e^(−((iπ)/6)) )^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ{Res(ϕ,e^((iπ)/6) )+Res(ϕ,−e^(−((iπ)/6)) )} Res(ϕ,e^((iπ)/6) )=lim_(z→e^((iπ)/6) ) (1/((2−1)!)){(z−e^((iπ)/6) )^2 ϕ(z)}^((1)) =lim_(z→e^((iπ)/6) ) {(1/((z+e^((iπ)/6) )(z^2 −e^(−((iπ)/3)) )^2 ))}^((1)) =lim_(z→e^((iπ)/6) ) −(((z^2 −e^(−((iπ)/3)) )^2 +(z+e^((iπ)/6) )×(4z)(z^2 −e^(−((iπ)/3)) ))/((z+e^((iπ)/6) )^2 (z^2 −e^(−((iπ)/3)) )^4 )) =−lim_(z→e^((iπ)/6) ) (((z^2 −e^(−((iπ)/3)) )+4z(z+e^((iπ)/6) ))/((z+e^((iπ)/6) )^2 (z^2 −e^(−((iπ)/3)) )^3 )) =−((e^((iπ)/3) −e^(−((iπ)/3)) +4e^((iπ)/6) ×2e^((iπ)/6) )/((2e^((iπ)/6) )^2 (e^((iπ)/3) −e^(−((iπ)/3)) )^3 )) =−((2isin((π/3))+8 e^((iπ)/3) )/(2e^((iπ)/3) (2i)^3 sin^3 ((π/3)))) =−((isin((π/3))+4e^((iπ)/3) )/(−8i e^((iπ)/3) sin^3 ((π/3)))) =e^(−((iπ)/3)) ×((sin((π/3))−4ie^((iπ)/3) )/(sin^3 ((π/3)))) =((sin((π/3))e^(−((iπ)/3)) −4i)/(8(((√3)/2))^3 )) =((((√3)/2)e^(−((iπ)/3)) −4i)/(3(√3))) ....be continued...

$${let}\:{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{let}\:\varphi\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{4}} −{z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{4}} −{z}^{\mathrm{2}} \:+\mathrm{1}=\mathrm{0}\Rightarrow{t}^{\mathrm{2}} −{t}+\mathrm{1}=\mathrm{0}\:\:\:\left({t}={z}^{\mathrm{2}} \right) \\ $$$$\Delta=\mathrm{1}−\mathrm{4}=−\mathrm{3}\:\Rightarrow{t}_{\mathrm{1}} =\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={e}^{\frac{{i}\pi}{\mathrm{3}}} \:{and}\:{t}_{\mathrm{2}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}={e}^{−\frac{{i}\pi}{\mathrm{3}}} \\ $$$${t}^{\mathrm{2}} −{t}+\mathrm{1}\:=\left({t}−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({t}−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\:\Rightarrow{z}^{\mathrm{4}} −{z}^{\mathrm{2}} \:\:+\mathrm{1}\:=\left({z}^{\mathrm{2}} −{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}^{\mathrm{2}} −{e}^{\frac{−{i}\pi}{\mathrm{3}}} \right) \\ $$$$=\left({z}−{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\left({z}+{e}^{\frac{−{i}\pi}{\mathrm{6}}} \right)\Rightarrow \\ $$$$\varphi\left({z}\right)=\frac{\mathrm{1}}{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)^{\mathrm{2}} \left({z}+{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)^{\mathrm{2}} \left({z}−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)^{\mathrm{2}} \left({z}+{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)^{\mathrm{2}} } \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\left\{{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\right\} \\ $$$${Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{6}}} } \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{6}}} } \:\:\:\left\{\frac{\mathrm{1}}{\left({z}+{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{6}}} } \:\:−\frac{\left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} +\left({z}+{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)×\left(\mathrm{4}{z}\right)\left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)}{\left({z}+{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)^{\mathrm{2}} \left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{4}} } \\ $$$$=−{lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{6}}} } \:\:\:\:\frac{\left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)+\mathrm{4}{z}\left({z}+{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)}{\left({z}+{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)^{\mathrm{2}} \left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{3}} } \\ $$$$=−\frac{{e}^{\frac{{i}\pi}{\mathrm{3}}} −{e}^{−\frac{{i}\pi}{\mathrm{3}}} +\mathrm{4}{e}^{\frac{{i}\pi}{\mathrm{6}}} ×\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{6}}} }{\left(\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)^{\mathrm{2}} \left({e}^{\frac{{i}\pi}{\mathrm{3}}} −{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{3}} }\:=−\frac{\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{3}}\right)+\mathrm{8}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} }{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{3}}} \left(\mathrm{2}{i}\right)^{\mathrm{3}} {sin}^{\mathrm{3}} \left(\frac{\pi}{\mathrm{3}}\right)} \\ $$$$=−\frac{{isin}\left(\frac{\pi}{\mathrm{3}}\right)+\mathrm{4}{e}^{\frac{{i}\pi}{\mathrm{3}}} }{−\mathrm{8}{i}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \:{sin}^{\mathrm{3}} \left(\frac{\pi}{\mathrm{3}}\right)}\:={e}^{−\frac{{i}\pi}{\mathrm{3}}} ×\frac{{sin}\left(\frac{\pi}{\mathrm{3}}\right)−\mathrm{4}{ie}^{\frac{{i}\pi}{\mathrm{3}}} }{{sin}^{\mathrm{3}} \left(\frac{\pi}{\mathrm{3}}\right)} \\ $$$$=\frac{{sin}\left(\frac{\pi}{\mathrm{3}}\right){e}^{−\frac{{i}\pi}{\mathrm{3}}} \:−\mathrm{4}{i}}{\mathrm{8}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} }\:=\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{3}}} −\mathrm{4}{i}}{\mathrm{3}\sqrt{\mathrm{3}}}\:\:....{be}\:{continued}... \\ $$

Commented by mathmax by abdo last updated on 01/Mar/20

pareametric method let ϕ(a) =∫_0 ^∞ (dx/(x^4 −x^2 +a)) witha>(1/4) we have ϕ^′ (a) =−∫_0 ^∞ (dx/((x^4 −x^2 +a)^2 )) ⇒∫_0 ^∞ (dx/((x^4 −x^2 +a)^2 ))=−ϕ^′ (a) we have 2ϕ(a) =∫_(−∞) ^(+∞) (dx/(x^4 −x^2 +a)) let W(z)=(1/(z^4 −z^2 +a)) poles of W? z^4 −z^2 +a =0 ⇒t^2 −t +a =0 (t=z^2 ) Δ=1−4a<0 ⇒Δ=(i(√(4a−1)))^2 ⇒z_1 =((1+i(√(4a−1)))/2) z_2 =((1−i(√(4a−1)))/2) ⇒W(z) =(1/((z^2 −z_1 )(z^2 −z_2 ))) =(1/((z−(√z_1 ))(z+(√z_1 ))(z−(√z_2 ))(z+(√z_2 )))) ∫_(−∞) ^(+∞) W(z)dz =2iπ {Res(W,(√z_1 )) +Res(W,−(√z_2 ))} Res(W,(√z_1 )) =(1/(2(√z_1 )(z_1 ^2 −z_2 ))) Res(W,−(√z_2 )) =(1/(−2(√z_2 )(z_2 ^2 −z_1 ))) =−(1/(2(√z_2 )(z_2 ^2 −z_1 ))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =iπ{ (1/((√z_1 )(z_1 ^2 −z_2 )))−(1/((√z_2 )(z_2 ^2 −z_1 )))} ∣z_1 ∣=(1/2)(√(1+4a−1))=(√a) ⇒z_1 =(√a)e^(iarctan(√(4a−1))) z_2 =(√a)e^(−iarctan((√(4a−1)))) ⇒(√z_1 )=^4 (√a)e^((i/2)arctan((√(4a−1)))) z_1 ^2 −z_2 =a e^(2iarctan((√(4a−1)))) −(√a)e^(−iarctan((√(4a−1)))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =iπ{ (1/((^4 (√a))e^(iarctan((√(4a−1)))) (a e^(2iarctan((√(4a−1)))) −(√a)e^(−iarctan((√(4a−1)))) ))−(1/((^4 (√a))e^(−iarctan((√(4a−1)))) (ae^(−2iarctan((√(4a−1)))) −(√a)e^(iarctan((√(4a−1)))) )))} ...be continued...

$${pareametric}\:{method}\:{let}\:\varphi\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+{a}}\:\:{witha}>\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${we}\:{have}\:\varphi^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+{a}\right)^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+{a}\right)^{\mathrm{2}} }=−\varphi^{'} \left({a}\right) \\ $$$${we}\:{have}\:\mathrm{2}\varphi\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+{a}}\:{let}\:{W}\left({z}\right)=\frac{\mathrm{1}}{{z}^{\mathrm{4}} −{z}^{\mathrm{2}} \:+{a}} \\ $$$${poles}\:{of}\:{W}? \\ $$$${z}^{\mathrm{4}} −{z}^{\mathrm{2}} \:+{a}\:=\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} −{t}\:+{a}\:=\mathrm{0}\:\:\left({t}={z}^{\mathrm{2}} \right) \\ $$$$\Delta=\mathrm{1}−\mathrm{4}{a}<\mathrm{0}\:\Rightarrow\Delta=\left({i}\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)^{\mathrm{2}} \:\Rightarrow{z}_{\mathrm{1}} =\frac{\mathrm{1}+{i}\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}} \\ $$$${z}_{\mathrm{2}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}}\:\Rightarrow{W}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −{z}_{\mathrm{1}} \right)\left({z}^{\mathrm{2}} −{z}_{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\left({z}−\sqrt{{z}_{\mathrm{1}} }\right)\left({z}+\sqrt{{z}_{\mathrm{1}} }\right)\left({z}−\sqrt{{z}_{\mathrm{2}} }\right)\left({z}+\sqrt{{z}_{\mathrm{2}} }\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left({W},\sqrt{{z}_{\mathrm{1}} }\right)\:+{Res}\left({W},−\sqrt{{z}_{\mathrm{2}} }\right)\right\} \\ $$$${Res}\left({W},\sqrt{{z}_{\mathrm{1}} }\right)\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{z}_{\mathrm{1}} }\left({z}_{\mathrm{1}} ^{\mathrm{2}} −{z}_{\mathrm{2}} \right)} \\ $$$${Res}\left({W},−\sqrt{{z}_{\mathrm{2}} }\right)\:=\frac{\mathrm{1}}{−\mathrm{2}\sqrt{{z}_{\mathrm{2}} }\left({z}_{\mathrm{2}} ^{\mathrm{2}} \:−{z}_{\mathrm{1}} \right)}\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{z}_{\mathrm{2}} }\left({z}_{\mathrm{2}} ^{\mathrm{2}} −{z}_{\mathrm{1}} \right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:={i}\pi\left\{\:\frac{\mathrm{1}}{\sqrt{{z}_{\mathrm{1}} }\left({z}_{\mathrm{1}} ^{\mathrm{2}} −{z}_{\mathrm{2}} \right)}−\frac{\mathrm{1}}{\sqrt{{z}_{\mathrm{2}} }\left({z}_{\mathrm{2}} ^{\mathrm{2}} −{z}_{\mathrm{1}} \right)}\right\} \\ $$$$\mid{z}_{\mathrm{1}} \mid=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+\mathrm{4}{a}−\mathrm{1}}=\sqrt{{a}}\:\Rightarrow{z}_{\mathrm{1}} =\sqrt{{a}}{e}^{{iarctan}\sqrt{\mathrm{4}{a}−\mathrm{1}}} \\ $$$${z}_{\mathrm{2}} =\sqrt{{a}}{e}^{−{iarctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} \:\Rightarrow\sqrt{{z}_{\mathrm{1}} }=^{\mathrm{4}} \sqrt{{a}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} \\ $$$${z}_{\mathrm{1}} ^{\mathrm{2}} −{z}_{\mathrm{2}} ={a}\:{e}^{\mathrm{2}{iarctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} −\sqrt{{a}}{e}^{−{iarctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:={i}\pi\left\{\:\frac{\mathrm{1}}{\left(^{\mathrm{4}} \sqrt{{a}}\right){e}^{{iarctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} \left({a}\:{e}^{\mathrm{2}{iarctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} −\sqrt{{a}}{e}^{−{iarctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} \right.}−\frac{\mathrm{1}}{\left(^{\mathrm{4}} \sqrt{{a}}\right){e}^{−{iarctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} \left({ae}^{−\mathrm{2}{iarctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} −\sqrt{{a}}{e}^{{iarctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} \right)}\right\} \\ $$$$...{be}\:{continued}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com